Two Sum

Contents

1. 1. Solution:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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Solution:

1.俩层遍历 暴力穷举
时间复杂度 O(N2)
空间复杂度 O(1)

public static int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[j] == target - nums[i]) {
                return new int[] { i, j };
            }
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

2.hashmap
时间复杂度 O(n)
空间复杂度 O(n)

public static int[] twoSum2(int[] nums, int target){
    Map<Integer, Integer> map = new HashMap<>();
    for(int i= 0;i<nums.length;i++){
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }

    }
     throw new IllegalArgumentException("No two sum solution");
}

3.改进的HashMap
时间复杂度 O(n)
空间复杂度 O(n)

public int[] twoSum3(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}